Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 190644 by cherokeesay last updated on 08/Apr/23

	Answered by a.lgnaoui last updated on 09/Apr/23
	$Algebrique solution CercleC_1 B:origine du Ref B(0,0) et de rayon 1 C_1 (B,1) ( B,E):C_1 ∩C_2 CercleC_2 (A,(1/2)) d origine A(−1,+(1/2)) equation cartesienne: { ((x^2 +y^2 =1 (1))),(((x+1)^2 +(y−(1/2))^2 =(1/4) (2))) :} ⇒ y=2(x+1) 5x^2 +8x+3=0 ⇒x=(−(3/5), −1) ⇒AM=(3/5) ou BM=(2/5) B(−1,0) △BEM BE=a AM=(3/5) BN=y=(√(1−(9/(25)))) =(4/5) BE^2 =BM^2 +ME^2 =(4/(25))+((16)/(25))=((20)/(25)) ⇒ a=((2(√5))/5) △EFC EC^2 =EF^2 +FC^2 FC=1−(4/5)=(1/5) EF=(3/5) EC^2 = (9/(25))+(1/(25))=((10)/(25)) ⇒ b=((√(10))/5) Rapport (a/b) =((2(√5))/( (√(10)))) donc (a/b)=(√2)$
	$Algebrique solution$ ${CercleC}_{1} B : origine du Ref B (0, 0)$ $et de rayon 1 C_{1} (B, 1) (B, E) : C_{1} \cap C_{2}$ ${CercleC}_{2} (A, \frac{1}{2}) d origine A (- 1, + \frac{1}{2})$ $equation cartesienne :$ ${\begin{cases} x^{2} + y^{2} = 1 (1) \\ {(x + 1)}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{4} (2) \end{cases}$ $\Rightarrow y = 2 (x + 1)$ ${5 x}^{2} + 8 x + 3 = 0 \Rightarrow x = (- \frac{3}{5}, - 1)$ $\Rightarrow AM = \frac{3}{5} ou BM = \frac{2}{5} B (- 1, 0)$ $△ BEM BE = a AM = \frac{3}{5}$ $BN = y = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ ${BE}^{2} = {BM}^{2} + {ME}^{2} = \frac{4}{25} + \frac{16}{25} = \frac{20}{25}$ $\Rightarrow a = \frac{2 \sqrt{5}}{5}$ $△ EFC {EC}^{2} = {EF}^{2} + {FC}^{2}$ $FC = 1 - \frac{4}{5} = \frac{1}{5} EF = \frac{3}{5}$ ${EC}^{2} = \frac{9}{25} + \frac{1}{25} = \frac{10}{25} \Rightarrow$ $b = \frac{\sqrt{10}}{5}$ $Rapport \frac{a}{b} = \frac{2 \sqrt{5}}{\sqrt{10}}$ $donc \frac{a}{b} = \sqrt{2}$
	Commented by a.lgnaoui last updated on 09/Apr/23

	Commented by cherokeesay last updated on 09/Apr/23

	$N i c e! t h a n k y o u s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum